I should prove if $\Sigma$ is a total order, but I have no idea how I could do that, can it involve the fact that element $0 \in \mathbb N$ doesn't belong to this relation because $x\Sigma 0$ would mean $\{0 \text{ is a divisor of } x\}$ which is not possible? And my gut feeling tells me that $\max(\mathbb N, \Sigma) = 1$ and $\nexists\min(\mathbb N, \Sigma)$ but I don't know if I am right. Will you please give me a nudge in the right direction?

@martini yes, I should have written $D(x)=\{y \in \mathbb N : y \mid x\}$! In regard to the misuse of $\land$ I thought it would connect two propositions so what I meant (paraphrasing from the ani-symmetry section) is "if $y \in \mathbb N$ are divisors of $x$ and $x \in \mathbb N$ are divisors of $y$ then...". My teacher has used this kind of notation in class so maybe I've wrongly assumed it is correct.
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haunted85Jul 12 '12 at 9:48

2 Answers
2

For $\Sigma$ to be a total order every two elements need to be comparable.

However $D(2)=\{1,2\}$ and $D(3)=\{1,3\}$ so neither is a subset of the other. Therefore $\lnot(2\Sigma3\lor 3\Sigma2)$ and so this is not a total order.

For $\min$ and $\max$, as you noted every number is a divisor of $0$; while no number except $1$ is a divisor of $1$, so $D(1)=\{1\}$ and $D(0)=\mathbb N$. It remains to show that $1\Sigma x$ for all $x$, but $1$ divides every number so it holds.